I search for the inverse Fourier transform $FT^{-1}$ of the 2D function $$f(u,v) = \frac{i u e^{-u^2/2 -v^2/2}}{u^2 + v^2}$$
with $u,v,x,y\in\mathbb{R}$and using the convention (but won't insist on)
$$FT^{-1}[f](x,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(u,v)e^{i (xu+yw)} du dv$$
I came across the problem in trying to solve a certain 2D Laplace equation ($\phi_{xx}+\phi_{yy}=-xe^{-x^2/2-y^2/2}$).
I have attempted two strategies, the one is transforming the problem into polar coordinates
$$ f(r,\theta)=\frac{i \cos{(\theta)} e^{-r^2/2}}{r} $$ and trying
$$ FT^{-1}[f(r,\theta)](s,\omega) = \frac{1}{\sqrt{2\pi}} \int_0^\infty \int_{-\pi}^{\pi} f(r,\theta)e^{i r s \cos{(\theta-\omega)}}rdrd\theta $$
which gave me (using CAS(MMA)):
$$ FT^{-1}[i \cos{(\theta)} e^{-r^2/2}] = \frac{1}{2} e^{-\frac{1}{2} s^2} \int_{-\pi}^{\pi} \cos (\theta) e^{\cos ^2(\theta-\omega)} \left(\text{erfi}\left(\frac{s \cos (\theta-\omega)}{\sqrt{2}}\right)+i\right) d\theta $$
Here I am stuck, (MMA says the calc is too long for my subscription).
The other try I gave it was to factorize the function into smaller parts and FT the factors and convolve the results:
$$FT^{-1}[\frac{i u e^{-u^2/2 -v^2/2}}{u^2 + v^2}] = FT^{-1}[\frac{i u}{u^2 + v^2}]*FT^{-1}[e^{-u^2/2 -v^2/2}]= (x / (x^2 + y^2)) * (e^{-x^2/2 -y^2/2}) $$
(where "*" shall denote "convolution"). Here I fail to perform the convolution (again, MMA says the computation takes too long for my subscription).
I am not sure of there exists an analytical/closed form solution. In the worst case, if it doesn't I would be also happy with a reasonable approximation.
Any help appreciated.
Here is a solution using Wirtinger derivatives. Let's change the variables $(x,y)$ to $(z, \bar{z})=(x+iy, x-iy)$; then $\partial_x=\partial_z + \partial_{\bar{z}}$ and $\partial_y=i(\partial_z-\partial_{\bar{z}})$, and the 2D Poisson equation $$ \phi_{xx}+\phi_{yy}=-xe^{-\frac{1}{2}(x^2+y^2)} \tag{1} $$ becomes $$ 4\phi_{z\bar{z}}=-\frac{1}{2}(z+\bar{z})e^{-\frac{1}{2}z\bar{z}}. \tag{2} $$ Integrating $(2)$ with respect to $z$ yields $$ \phi_{\bar z}=\frac{1}{4}\left(1+\frac{z}{\bar z}+\frac{2}{\bar{z}^2}\right) e^{-\frac{1}{2}z\bar{z}} \tag{3} $$ (plus an arbitrary function of $\bar{z}$, which we shall ignore because we want to find only a particular solution to the PDE $(1)$). Integrating $(3)$ with respect to $\bar{z}$ we finally obtain $$ \phi = -\frac{1}{2}\left(\frac{1}{z}+\frac{1}{\bar{z}}\right) e^{-\frac{1}{2}z\bar{z}}=-\frac{x}{x^2+y^2}e^{-\frac{1}{2}(x^2+y^2)}. \tag{4} $$